Complex Analysis - Maths KU

362 Chapter 6 Series and Residues
z = –1
C r
y
A
D
C R
B
E
Figure 6.15 Contour for Example 6
x
EXAMPLE 6 Integration along a Branch Cut
� ∞
1
Evaluate √ dx.
x(x +1)
0
Solution First observe that the real integral is improper for two reasons.
Notice an infinite discontinuity at x = 0 and the infinite limit of integration.
Moreover, it can be argued from the facts that the integrand behaves like
x−1/2 near the origin and like x−3/2 as x →∞, that the integral converges.
�
1
We form the integral
C z1/2 dz, where C is the closed contour
(z +1)
shown in Figure 6.15 consisting of four components: Cr and CR are portions
of circles, and AB and ED are parallel horizontal line segments running along
opposite sides of the branch cut.The integrand f(z) of the contour integral
is single valued and analytic on and within C, except for the simple pole at
z = −1 =eπi .Hence we can write
�
1
C z1/2 dz =2πi Res(f(z), −1)
(z +1)
� � � �
or
+ + + =2πi Res(f(z), −1). (23)
CR ED Cr AB
Despite what is shown in Figure 6.15, it is permissible to think that the line
segments AB and ED actually rest on the positive real axis, more precisely,
AB coincides with the upper side of the positive real axis for which θ = 0 and
ED coincides with the lower side of the positive real axis for which θ =2π.
On AB, z = xe0i , and on ED, z = xe (0+2π)i = xe2πi , so that
� � r
(xe
=
ED R
2πi ) −1/2
xe2πi +1 (e2πi � r
x
dx) =−
R
−1/2 � R
dx =
x +1 r
�
x−1/2 dx (24)
x +1
� R
(xe
=
AB r
0i ) −1/2
xe0i +1 (e0i � R
x
dx) =
r
−1/2
dx.
x +1
(25)
and
Now with z = reiθ and z = Reiθ on Cr and CR, respectively, it can be shown,
by analysis similar to that given in Example 2 and in the proof of Theorem
6.17, that �
�
→ 0asr→ 0 and → 0asR→∞. Thus from (23), (24),
Cr CR
and (25) we see that
�� � � �
�
lim
r→0
R→∞ CR
+
ED
+
Cr
+
AB
=2πi Res(f(z), −1)
is the same as
� ∞
2
0
Finally, from (4) of Section 6.5,
Res(f(z), −1) = z −1/2
�
�
1
√ x(x +1) dx =2πi Res(f(z), −1). (26)
� z=e πi
= e −πi/2 = −i